2. Hitunglah (a) \( \int_{1}^{2}\left(\frac{1}{t}-i\right)^{2} d t \) dan (b) \( \int_{0}^{\pi / 2} 2 e^{i t} d t \).

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{1}^{2} \left(\frac{1}{t}-i\right)^{2} dt\) - step1: Evaluate the integral: \(\int \left(\frac{1}{t}-i\right)^{2} dt\) - step2: Rewrite the expression: \(\int \left(\frac{1}{t^{2}}-\frac{i}{t}-\frac{i}{t}-1\right) dt\) - step3: Use properties of integrals: \(\int \frac{1}{t^{2}} dt+\int -\frac{i}{t} dt+\int -\frac{i}{t} dt+\int -1 dt\) - step4: Evaluate the integral: \(\frac{t^{-1}}{-1}+\int -\frac{i}{t} dt+\int -\frac{i}{t} dt+\int -1 dt\) - step5: Evaluate the integral: \(\frac{t^{-1}}{-1}-i\ln{\left(\left|t\right|\right)}+\int -\frac{i}{t} dt+\int -1 dt\) - step6: Evaluate the integral: \(\frac{t^{-1}}{-1}-i\ln{\left(\left|t\right|\right)}-i\ln{\left(\left|t\right|\right)}+\int -1 dt\) - step7: Evaluate the integral: \(\frac{t^{-1}}{-1}-i\ln{\left(\left|t\right|\right)}-i\ln{\left(\left|t\right|\right)}-t\) - step8: Rewrite the expression: \(-\frac{1}{t}-i\ln{\left(\left|t\right|\right)}-i\ln{\left(\left|t\right|\right)}-t\) - step9: Subtract the terms: \(-\frac{1}{t}-2i\ln{\left(\left|t\right|\right)}-t\) - step10: Return the limits: \(\left(-\frac{1}{t}-2i\ln{\left(\left|t\right|\right)}-t\right)\bigg |_{1}^{2}\) - step11: Calculate the value: \(-\frac{1}{2}-2\ln{\left(2\right)}\times i\) Mari kita hitung kedua integral satu per satu. ### (a) Menghitung \( \int_{1}^{2}\left(\frac{1}{t}-i\right)^{2} d t \) Pertama, kita perlu menyederhanakan ekspresi di dalam integral: \[ \left(\frac{1}{t}-i\right)^{2} = \left(\frac{1}{t}\right)^{2} - 2\left(\frac{1}{t}\right)(i) + i^{2} = \frac{1}{t^{2}} - \frac{2i}{t} - 1 \] Sekarang kita substitusi ke dalam integral: \[ \int_{1}^{2}\left(\frac{1}{t}-i\right)^{2} d t = \int_{1}^{2} \left(\frac{1}{t^{2}} - \frac{2i}{t} - 1\right) d t \] Kita dapat memisahkan integral ini menjadi tiga bagian: \[ \int_{1}^{2} \frac{1}{t^{2}} d t - 2i \int_{1}^{2} \frac{1}{t} d t - \int_{1}^{2} 1 d t \] Mari kita hitung masing-masing integral: 1. \( \int_{1}^{2} \frac{1}{t^{2}} d t = \left[-\frac{1}{t}\right]_{1}^{2} = -\frac{1}{2} + 1 = \frac{1}{2} \) 2. \( \int_{1}^{2} \frac{1}{t} d t = [\ln(t)]_{1}^{2} = \ln(2) - \ln(1) = \ln(2) \) 3. \( \int_{1}^{2} 1 d t = [t]_{1}^{2} = 2 - 1 = 1 \) Sekarang kita substitusi hasil-hasil ini ke dalam integral: \[ \int_{1}^{2}\left(\frac{1}{t}-i\right)^{2} d t = \frac{1}{2} - 2i \ln(2) - 1 \] Menyederhanakan hasil: \[ = -\frac{1}{2} - 2i \ln(2) \] ### (b) Menghitung \( \int_{0}^{\pi / 2} 2 e^{i t} d t \) Kita dapat menghitung integral ini dengan menggunakan rumus integral dari fungsi eksponensial kompleks: \[ \int e^{i t} d t = \frac{1}{i} e^{i t} + C \] Maka, \[ \int 2 e^{i t} d t = 2 \cdot \frac{1}{i} e^{i t} = \frac{2}{i} e^{i t} \] Sekarang kita evaluasi dari \(0\) sampai \(\frac{\pi}{2}\): \[ \left[ \frac{2}{i} e^{i t} \right]_{0}^{\frac{\pi}{2}} = \frac{2}{i} \left( e^{i \frac{\pi}{2}} - e^{0} \right) = \frac{2}{i} \left( i - 1 \right) \] Menyederhanakan hasil: \[ = \frac{2(i - 1)}{i} = 2 - 2i \] ### Kesimpulan Hasil dari kedua integral adalah: (a) \( -\frac{1}{2} - 2i \ln(2) \) (b) \( 2 - 2i \)